A 1.00-kg glider attached to a spring with a force constant 25.0 N/m oscillates on a frictionless, horizontal air track. At t = 0, the glider is released from rest at x = -2.80 cm (that is, the spring is compressed by 2.80 cm). (a) Find the period of the glider's motion. How does the period depend on the mass and the spring constant? Does it depend on the amplitude of oscillation? s (b) Find the maximum values of its speed and acceleration. speed m/s acceleration m/s2 (c) Find the position, velocity, and acceleration as functions of time. (Where position is in m, velocity is in m/s, acceleration is in m/s2, and t is in s. Use the following as necessary: t.) x(t) = v(t) = a(t) =

a)  T = 1.26 s , b)  v_max = 0.14 m / s ,  a_max = 0.7 m / s²

c) x = 0.028 cos (5 t) ,    v = - 0.14 sin 5t,   a = - 0.7 cos 5t

Explanation:

This is a simple harmonic motion exercise that is described by the equation

    x = A cos (wt +Ф)

with

          w = √ (k / m)

let's apply this expression to our case

a) Angular velocity is related to frequency

          w = 2π f

frequency and period are related

          f = 1 / T

we substitute

         2π / T = √ (k / m)

         T = 2π √(m / k)

let's calculate

         T = 2π √(1/25)

          T = 1.26 s

In the expression for the period, the amplitude does not appear, therefore there is no dependence, as long as Hooke's law is fulfilled, which is correct for small amplitudes.

b) in the initial equation we have the position as a function of time, let's use the definition of speed and acceleration

           v = dx / dt

           v = - A w sin (wt + Ф)

the speed is maximum when the sine is -1

            v_max = A w

            w = √ (k / m)

            w = √ 25/1

            w = 5 rad / s

the amplitude of the movement is equal to the maximum compression of the spring

            A = 2.8 cm = 0.028 m

we substitute

            v_max = 0.028 5

            v_max = 0.14 m / s

acceleration

             a = dv / dt

             a = - A w² cos (wt + Ф)

the acceleration is maximum when the cosine is -1

             a_max = A w²

let's calculate

             a_max = 0.028 5²

             a_max = 0.7 m / s²

c) let's start by finding the phase constant

              v = -A w cos (wt + Ф)

at t = 0 they indicate that the system has v = 0

              0 = -A w sin (0 + Ф)

              Ф = sin⁻¹ 0

              Ф = 0

we write the equation

            x = 0.028 cos (5 t)

           v = - A w sin (wt + Ф)

           v = - 0.028 5 sin (5t + 0)

           v = - 0.14 sin 5t

acceleration

           a = - A w² cos (wt + Ф)

           a = - 0.028 5 2 cos (5 t + 0)

           a = - 0.7 cos 5t

answer: b.

explanation:

the closer the planet is to the sun, the larger the gravity